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hey everyone welcome back to the channel i hope you guys are doing extremely well so today we will be solving the problem recover a binary search tree now what does the problem state the problem states that you will be given the root of a binary search tree and in that that root will be representing a complete vst now you would be knowing what is a pst a bst is a binary tree where you have the node and everything to the left of that node is smaller than it and everything to the right of that node is greater than it that is uh what is the property of bst now what is the question states the question states that whatever bst is given to you in that there are two nodes which are swapped okay now your job is to correct the bst for an example if i say that okay is this a valid bst the answer to that will be no why because this is three and everything to the right of it should be greater than 3 but there is something as 2 which is smaller than 3 so if i if i do one thing like if i just take this and if i take this and i swap now is this a valid bst yes this is a valid bst so what we did find is you're given a bst where like it's already mentioned that there will always be two nodes swapped remember this there will always be a case where two nodes will be swapped so over here two notes are not more than not lesser than that two notes will be swapped so we saw that the notes that were swapped towards this and this so the moment i swapped the values now this is indeed a valid vst so the question is very straightforward you'll be given a route which represents a bsd uh but that bst will have two nodes trapped so if to correct the bst and return basically just correct the bst you don't have to return it because if you just correct it that should be okay so basically make sure you change the swap the notes so so how do you solve this problem what is the extreme brute uh method that comes to your brain the extreme root method that comes to my brain is one thing i know for sure is if i do a inorder traversal if i do an inorder traversal for the given bst then the inorder will not be sorted because there are two nodes which are swapped so what i can do is i will do an in-order traversal yes any traversal uh for that reason like you can do any traversal it can be pre-order post order just make sure you sort that traversal yes just make sure you sort that travis once you sort that traversal it will be the correct inorder of...